Find the power of $Is_1$ using circuit reduction methods.

Solution
$R_1$ and $R_4$ are parallel. $R_2$ and $R_3$ are also parallel. Therefore:

where $R_{14} = R_1 || R_4 = 5 \Omega$ and $R_{23} = R_2 || R_3 = 3 \Omega$. Now, $R_{14}$ and $R_{23}$ are in series. Hence,

and $R_{1234} = R_{14} + R_{23} = 8 \Omega$. Voltage across the current source, i.e. $V_{Is_1}$ can be found by the resistor:
$V_{Is_1} = Is_1 \times R_{1234} = 80 v$. Thus,

$P_{Is_1} = - V_{Is_1} \times Is_1 = -800$ , supplying.